PHYSICS 231Review problems for midterm 2

Topic 5: Energy and Work and Power

Topic 6: Momentum and Collisions

Topic 7: Oscillations (spring and pendulum)

Topic 8: Rotational Motion

• The 2nd exam will be Wednesday October 28.

• The exam will take place right here in BPS 1410.

• We will have assigned seating, so show up early.

• You need to show up 10 min early to guarantee your seat.

• If you are sick, you MUST have a note from a doctor on the doctor’s letterhead or prescription pad.

• You CANNOT use cell phones during the exam.

• You can (should!) bring two hand written equation sheets.up to two sides each on a 8.5x11 sheet of paper

2

atvv f 0

221 attvx o

221 attvx f

tvtvvx f )( 021

)( 20

221 vvx fa

(A)

(B)

(C)

(D)

(E)

Energy

Potential energy (PE) Energy associated with position.

Gravitational PE: mgh Energy associated with position in grav. field.

Kinetic energy KE: ½mv2 Energy associated with motion

Elastic PE: ½kx2 energy stored in stretched/compressed spring.

Mechanical energy ME: ME = KE + PE

Conservative forces MEi – MEf = 0

Non-conservative forces MEi - MEf = Enc

Work (the amount of energy transfer)

W=(Fcos) x 

To measure how fast we transfer the energy we define:Power: The rate of energy transfer

Power = P = (Work/time) = (W/t)   (J/s=Watt) 

P = (Fcos)·x/t = (Fcos)·vaverage

Equations of motion

Linear motion Angular motion

x(t) = xo + vot + ½at2

v(t) = vo + at

(t) = o + ot + ½t2

(t) = o + t

Angular motion = Rotational Motion!

Momentum and ImpulseF = m a Newton’s 2nd lawF = m v/t a=v/tF = m (vfinal-vinital)/t

Define p = mv p: momentum (kg m/s)F= (pfinal-pinitial)/t

F= p/tDefinition: p = Impulse

CONSERVATION OF MOMENTUM (for a closed system)

m1v1f + m2v2f = m1v1i + m2v2ior p1i + p2i = p1f + p2f

addition of vectors example pi = p1f + p2f

Inelastic Solution

vf = [m1 v1i + m2v2i] / (m1+m2)

With m2 = c m1

vf = [v1i + cv2i] / (1+c)

Special case m2 = m1 (c=1)

vf = [v1i + v2i] / 2

Conservation of momentum: m1v1i + m2v2i = m1v1f + m2v2f

After the collision m1 and m2 form one new object with massM = m1 + m2 and one velocity vf =v1f = v2f

m1v1i + m2v2i = vf (m1 + m2)

KEi - KEf = Enc

Elastic Solution

v1f = [(m1 –m2) v1i + 2m2v2i] / (m1+m2)

v2f = [(m2 –m1) v2i + 2m1v1i] / (m1+m2)

For m2 = c m1

v1f = [(1-c) v1i + 2cv2i] / (1+c)

v2f = [(c-1) v2i + 2v1i] / (1+c)

Special case m2 = m1 (c=1)

v1f = v2i

v2f = v1i

KEi - KEf = 0

SpringTotal ME at any displacement x: ½mv2 + ½kx2

Total ME at max. displacement A: ½kA2

Conservation of ME: ½kA2 = ½mv2 + ½kx2

So: v=±[(A2-x2)k/m]

position X velocity V acceleration

+A 0 -kA/m

0 ±A(k/m) 0

-A 0 kA/m

A

-A

velocity v

x x(t)=Acos(t)

v(t)=-Asin(t)

a(t)=-2Acos(t)

=(k/m)

A(k/m)

-A(k/m)

-kA/m

a

kA/m

time (s)

time (s)

time (s)

pendulum vs spring

parameter spring pendulum

restoring force F

F=-kx F=-(mg/L)s

angular frequency

=(k/m) =(g/L)

Lg

mLmg

mk

_/

the pendulum

turnonefor time period T

radiusr

radiansin with arclength rs

velocityntial tange rt

rtsv

onaccelerati lcentripeta)/( 22 rrvac

(rad/s)in locity angular ve22 fTt

circular motion

on acceleratiangular t

onaccelerati ial tangent rt

rtva

An ExampleA

B

5m

6m

The angular velocity of A is 2 rad/s. 1) What is its tangential velocity?

If B is keeping pace with A,2) What is its angular velocity?3) What is its tangential velocity?

1) v = r = 26 = 12 m/s

12 m/s

2) Must be the same: 2 rad/s

3) v = r = 25 = 10 m/s

10 m/s

Translational equilibrium: F=ma=0 The center of gravitydoes not move!

Rotational equilibrium: =0 The object does notrotate

Mechanical equilibrium: F=ma=0 & =0 No movement!

Torque: =Fd

ii

iii

CG m

xmx

ii

iii

CG m

ymyCenter of

Gravity:

=IMoment of inertia I:I=(miri

2)

I = mr2 all mass is at the same distance r

compare with: F=maThe moment of inertiain rotations is similar tothe mass in Newton’s 2nd law.

m1

m2

m3r1

r2r3

Moment of inertia

Rotational kinetic energy

Consider a object rotatingwith constant velocity. Each pointmoves with velocity vi. The totalkinetic energy is:

Irmrmvmi i

iiiiii

ii222222

21

21

21

21

KEr=½I2

Conservation of energy for rotating object must include:rotational and translational kinetic energy and potential energy

[PE+KEt+KEr]initial= [PE+KEt+KEr]final

miri

v

Which one goes fastestUse conservation of mechanical energy.At top (at rest): [½mv2+mgh+½I2] = mghAt bottom:

[½mv2 + ½I2] = [½mv2 + ½(kmr2)v2/r2](used: I=kmr2 and =v/r)

Simplify: [½mv2+½kmv2]=(1+k)½mv2

Combine:Etop=Ebottommgh=(1+k)½mv2

v=[2gh/(1+k)]No mass dependence!!

object I k

A Cylindrical shell

Mr2 1

B Solid cylinder (1/2) mr2 0.5C Thin spherical

shell(2/3) mr2 0.666

D Solid sphere (2/5) mr2 0.4

Angular momentum

0L then 0 if

0

00

tL

tLLIL

tII

tII

Conservation of angular momentumIf the net torque equals zero, theangular momentum L does not change

Iii = Iff